Question

${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+......+{\tan }^{-1}\left(\frac{1}{n^2+n+1}\right)$ equals ?

• A. ${\tan }^{-1}\left(\frac{1}{3n}\right)$
• B. ${\tan }^{-1}\left(\frac{1}{2n+1}\right)$
• C. ${\tan }^{-1}\left(\frac{n}{n+2}\right)$
• D. ${\tan }^{-1}\left(\frac{1}{n+2}\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{n}{n+2}\right)$

The $r$th term of the equation is

${\tan }^{-1}\left(\frac{1}{r^2+r+1}\right)={\tan }^{-1}\left(\frac{r+1-r}{r^2+r+1}\right)={\tan }^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)={\tan }^{-1}(r+1)-{\tan }^{-1}r$

So summation is:

$\sum _{r=1}^n{\tan }^{-1}\left(\frac{1}{r^2+r+1}\right)$

$=({\tan }^{-1}3-{\tan }^{-1}2)+({\tan }^{-1}4-{\tan }^{-1}3)$

$={\tan }^{-1}(n+1)-{\tan }^{-1}1={\tan }^{-1}\left(\frac{n+1-1}{1+(n+1)\left(1\right)}\right)={\tan }^{-1}\left(\frac{n}{n+2}\right)$